回到作案现场：http://acdream.info/onecontest/1014
前言：自己出份山寨版的解题报告。
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A - Boy or Girl
统计用了多少种单词。分奇偶数输出结论。能够用STL的set实现。我对map比較熟，所以果断用map水了。
/*
* this code is made by code4101
* Problem: 1080
* Verdict: Accepted
* Submission Date: 2014-05-19 01:47:38
* Time: 0 MS
* Memory: 1676 KB
*/
#include <iostream>
#include <map>
using namespace std;
int main()
{
string s;
map<char, int> a;
while (cin >> s)
{
a.clear();
for (int i = 0; i < s.size(); i++) a[s[i]]++;
if (a.size()&1) cout << "IGNORE HIM!\n";
else cout << "CHAT WITH HER!\n";
}
return 0;
}
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B - Walking in the Rain
题目非常多无用的干扰信息。题意非常easy。就是给n个砖块顺序排好。告诉你每块砖几天后坏。然后问几天后"路径不通"。不通的定义是：①第一个或最后一个坏了 ②存在连续的两个坏了。
我是用递归。每次找到最小的非0值模拟暴力解（比赛中的我常常该去吃药了）。后来__M子__给的思想是一次遍历就够了：遍历max（a[i]，a[i+1]），记录最小值。再与头尾比較。
/*
* this code is made by code4101
* Problem: 1079
* Verdict: Accepted
* Submission Date: 2014-05-19 02:01:47
* Time: 0 MS
* Memory: 1672 KB
*/
#include <iostream>
using namespace std;
int main()
{
int n, a[1010], i, ans, t;
while (cin >> n)
{
for (i = 0; i < n; i++) cin >> a[i];
ans = min(a[0], a[n-1]);
for (i = 0; i < n-1; i++)
{
t = max(a[i], a[i+1]);
if (t < ans) ans = t;
}
cout << ans << "\n";
}
return 0;
}
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C - Cutting Figure
就是给一个连通图。让你用最小的删除操作把它变成非连通图。事实上答案范围一定在-1~4之间（随便找一个点，把上下左右全删除，这个点就孤立了，最坏情况下把其变为非连通图）。-1代表无法实现。再进一步分析。事实上答案仅仅有-1、1、2三种情况（角上的点最多去掉2个相邻的就孤立了）。
-1就是初始集合元素不超过2个，这样无论怎么删都是连通的。1的话要遍历，一个个顶点去掉，用BFS尝试看是不是非连通图（BFS用队列实现。这个不懂去补基础吧）。假设不是前两者，那么答案就一定是2了。
/*
* this code is made by code4101
* Problem: 1078
* Verdict: Accepted
* Submission Date: 2014-05-18 12:04:24
* Time: 0 MS
* Memory: 1688 KB
*/
#include <cstdio>
#include <iostream>
#include <cstring>
#include <queue>
using namespace std;
int n, m;
bool a[60][60], b[60][60], zero[60][60] = {};
struct point{
int x, y;
};
// 推断一个b是不是连通
bool isconnected(int x, int y)
{
// 先进行一遍BFS
queue<point> A;
point p, t;
p.x = x; p.y = y;
A.push(p); b[p.x][p.y] = 0;
while (A.size())
{
t = A.front(); A.pop();
if (b[p.x = t.x-1][p.y = t.y]) A.push(p), b[p.x][p.y] = 0;
if (b[p.x = t.x][p.y = t.y-1]) A.push(p), b[p.x][p.y] = 0;
if (b[p.x = t.x][p.y = t.y+1]) A.push(p), b[p.x][p.y] = 0;
if (b[p.x = t.x+1][p.y = t.y]) A.push(p), b[p.x][p.y] = 0;
}
if (memcmp(zero, b, sizeof(b))) return 0;
else return 1;
}
// 推断去掉随意一个#是不是连通
int fun()
{
int i, j;
for (i = 1; i <= n; i++)
{
for (j = 1; j <= m; j++) if (a[i][j])
{
memcpy(b, a, sizeof(a));
b[i][j] = 0;
int x, y; // 起点
if (b[x = i-1][y = j]);
else if (b[x = i][y = j-1]);
else if (b[x = i][y = j+1]);
else if (b[x = i+1][y = j]);
if (!isconnected(x, y)) return 1;
}
}
return 2;
}
int main()
{
while (cin >> n >> m)
{
memset(a, 0, sizeof(a));
int sum = 0, i, j;
for (i = 1; i <= n; i++)
{
getchar();
for (j = 1; j <= m; j++)
{
if (getchar() == '#')
{
sum += a[i][j] = 1;
}
}
}
if (sum <= 2) cout << "-1\n";
else cout << fun() << "\n";
}
return 0;
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D - LCM Challenge
这题我在蓝桥杯做过：http://www.coolnote.cn/problems/66/；这篇文章分析的很具体：http://blog.csdn.net/u011669700/article/details/18702757，事实上我到如今还是不太懂原理；原题来自这里：http://acm.hrbust.edu.cn/index.php?
m=ProblemSet&a=showProblem&problem_id=1632。
/*
* this code is made by code4101
* Problem: 1077
* Verdict: Accepted
* Submission Date: 2014-05-19 02:15:50
* Time: 0 MS
* Memory: 1672 KB
*/
#include <iostream>
using namespace std;
int main()
{
unsigned long long n, ans;
while (cin >> n)
{
if (n <= 2) ans = n;
else if (n & 1) ans = n * (n - 1) * (n - 2);
else if (n % 3) ans = n * (n - 1) * (n - 3);
else ans = (n - 1) * (n - 2) * (n - 3);
cout << ans << '\n';
}
return 0;
}
